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Brightness of distant galaxies problem

  1. Aug 29, 2011 #1
    1. The problem statement, all variables and given/known data

    An object gives out a spectral line of 919nm compared to that of 589nm in a laboratory. It has a brigtness of 1.2 x 10-11 W m-2. The objective is to find out the brightness of the object if it were at 700 Mpc.

    2. Relevant equations

    1. Z (redshift)= change in wavelength / original wavelength

    2. V= Z X C

    3. V= H0D

    4. F= L / 4 pi r2

    3. The attempt at a solution

    The redshift value I find to be 0.56 using equation 1. Using this I calculate the speed as 168000 km s-1 using equation 2. I then work out the distance of the object to be 2400Mpc using equation 3 rearranged. now from here do I use equation 4 to calculate the luminosity and then change the distance to 700Mpc or do I divide 2400Mpc by 700Mpc to get how many times closer the object will be square the answer ( 3.432) and times it by the original brightness 1.2 x 10-11 W m-2?
  2. jcsd
  3. Aug 30, 2011 #2


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    Try both ways and see if they give the same answer or not.
  4. Aug 31, 2011 #3
    I tried both, equation 4 gives me an answer of 4.932 x 10-9 and using divide and square i get 1.411 x 10-10 both seem plausible answers.
  5. Aug 31, 2011 #4


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    In what units?

    If they're the units I suspect you meant, one of those answers is wrong - and not just in the obvious sense that they can't both be right; you miscalculated one of them. Go back and check your math.
  6. Aug 31, 2011 #5
    both should be in units of W m-2. In equation 4 I converted Mpc to metres same luminosity is Watts. My equation for 4 looks like this rearranged.

    L= F x 4 pi r2

    Putting in the values gives 1.2 x 10-11 W m-2(F) x 4 pi (7.44x 1025)2 m (r2) = 8.35 x 1041 W (L)

    With that I get luminosity to be 8.35 x 1041 W (L)

    Using equation 4 as F= L / 4 pi r2 and using the new value for Luminosity I get.

    8.35 x 1041 W / (4 pi (2.17 x 1025)2)m2 = 4.932 x 10-9 W m-2

    Can anyone tell me where in that I am going wrong as the equations and units all seem to add up if anything I feel this may be the correct answer and the other method using 3.432 may be wrong. However as 3.432 does not have units is shouldnt effect W m-2.
  7. Aug 31, 2011 #6


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    This is the problem. The result of that calculation is not [itex]4.932\times 10^{-9}\frac{\mathrm{W}}{\mathrm{m}^2}[/itex].
  8. Aug 31, 2011 #7
    I just calculated it and got 1.411 x 10-10 I think the problem was I was squaring the entire bottom half of the divide rather than the distance. A stupid mistake but guess we all make them. I assume since both equations give the same answer that is the right answer?
  9. Aug 31, 2011 #8


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    Yeah, everybody makes stupid mistakes ;)

    Usually when you can do something two different ways and get the same answer, chances are good that you got it right.
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